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Question

我有一个熊猫数据框，看起来像这样：

employeeId     cumbId firstName lastName        emailAddress  \
0    E123456  102939485    Andrew   Hoover   hoovera@xyz.com   
1    E123457  675849302      Curt   Austin  austinc1@xyz.com   
2    E123458  354852739   Celeste  Riddick  riddickc@xyz.com   
3    E123459  937463528     Hazel   Tooley   tooleyh@xyz.com     

  employeeIdTypeCode cumbIDTypeCode entityCode sourceCode roleCode  
0                001            002      AE      AWB    EMPLR  
1                001            002      AE      AWB    EMPLR  
2                001            002      AE      AWB    EMPLR  
3                001            002      AE      AWB    EMPLR

我希望它对熊猫数据框中的每个ID和IDtypecode看起来像这样：

idvalue   IDTypeCode  firstName lastName  emailAddress  entityCode  sourceCode  roleCode  CodeName
E123456   001         Andrew    Hoover    hoovera@xyz.com AE        AWB         EMPLR     1
102939485 002         Andrew    Hoover    hoovera@xyz.com AE        AWB         EMPLR     1

这可以通过熊猫数据框中的某些功能来实现吗？我还希望它根据数据框中ID的数量是动态的。

动态的意思是，如果有3个Ids，那么它应该是这样的：

idvalue   IDTypeCode  firstName lastName  emailAddress  entityCode  sourceCode  roleCode  CodeName
A123456   001         Andrew    Hoover    hoovera@xyz.com AE        AWB         EMPLR     1
102939485 002         Andrew    Hoover    hoovera@xyz.com AE        AWB         EMPLR     1
M1000     003         Andrew    Hoover    hoovera@xyz.com AE        AWB         EMPLR     1

谢谢！

Answer 1

我认为这就是您要寻找的... 您可以在拆分数据框的各个部分后使用concat：

# create a new df without the id columns
df2 = df.loc[:, ~df.columns.isin(['employeeId','employeeIdTypeCode'])]

# rename columns to match the df columns names that they "match" to
df2 = df2.rename(columns={'cumbId':'employeeId', 'cumbIDTypeCode':'employeeIdTypeCode'})

# concat you dataframes
pd.concat([df,df2], sort=False).drop(columns=['cumbId','cumbIDTypeCode']).sort_values('firstName')

# rename columns here if you want

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# sample df
  employeeId     cumbId  otherId1 firstName lastName      emailAddress  \
0    E123456  102939485         5    Andrew   Hoover   hoovera@xyz.com   
1    E123457  675849302         5      Curt   Austin  austinc1@xyz.com   
2    E123458  354852739         5   Celeste  Riddick  riddickc@xyz.com   
3    E123459  937463528         5     Hazel   Tooley   tooleyh@xyz.com   

   employeeIdTypeCode  cumbIDTypeCode  otherIdTypeCode1 entityCode sourceCode  \
0                   1               2                 6         AE        AWB   
1                   1               2                 6         AE        AWB   
2                   1               2                 6         AE        AWB   
3                   1               2                 6         AE        AWB   

  roleCode  
0    EMPLR  
1    EMPLR  
2    EMPLR  
3    EMPLR

必须有一些规则：

规则1.总有两个“匹配列” 规则2.所有匹配的ID彼此相邻规则3.您知道Ids组的数量（要添加的行）

def myFunc(df, num_id): # num_id is the number of id groups 
    # find all columns that contain the string id
    id_col = df.loc[:, df.columns.str.lower().str.contains('id')].columns

    # rename columns to id_0 and id_1
    df = df.rename(columns=dict(zip(df.loc[:, df.columns.str.lower().str.contains('id')].columns,
                                ['id_'+str(i) for i in range(int(len(id_col)/num_id)) for x in range(num_id)])))

    # groupby columns and values.tolist
    new = df.groupby(df.columns.values, axis=1).agg(lambda x: x.values.tolist())

    data = []

    # for-loop to explode the lists
    for n in range(len(new.loc[:, new.columns.str.lower().str.contains('id')].columns)):
        s = new.loc[:, new.columns.str.lower().str.contains('id')]
        i = np.arange(len(new)).repeat(s.iloc[:,n].str.len())
        data.append(new.iloc[i, :-1].assign(**{'id_'+str(n): np.concatenate(s.iloc[:,n].values)}))

    # remove the list from all cells
    data0 = data[0].applymap(lambda x: x[0] if isinstance(x, list) else x).drop_duplicates()
    data1 = data[1].applymap(lambda x: x[0] if isinstance(x, list) else x).drop_duplicates()

    # update dataframes
    data0.update(data1[['id_1']])

    return data0

myFunc(df,3)


      emailAddress entityCode firstName       id_0  id_1 lastName roleCode
0   hoovera@xyz.com         AE    Andrew    E123456     1   Hoover    EMPLR
0   hoovera@xyz.com         AE    Andrew  102939485     2   Hoover    EMPLR
0   hoovera@xyz.com         AE    Andrew          5     6   Hoover    EMPLR
1  austinc1@xyz.com         AE      Curt    E123457     1   Austin    EMPLR
1  austinc1@xyz.com         AE      Curt  675849302     2   Austin    EMPLR
1  austinc1@xyz.com         AE      Curt          5     6   Austin    EMPLR
2  riddickc@xyz.com         AE   Celeste    E123458     1  Riddick    EMPLR
2  riddickc@xyz.com         AE   Celeste  354852739     2  Riddick    EMPLR
2  riddickc@xyz.com         AE   Celeste          5     6  Riddick    EMPLR
3   tooleyh@xyz.com         AE     Hazel    E123459     1   Tooley    EMPLR
3   tooleyh@xyz.com         AE     Hazel  937463528     2   Tooley    EMPLR
3   tooleyh@xyz.com         AE     Hazel          5     6   Tooley    EMPLR

Answer 2

据我了解，您要为每个源行生成2行：

employeeId（重命名为idvalue），然后IDTypeCode ='001'，然后是“ remainig”列（但不是全部），最后是CodeName ='1'。
cumbId，然后IDTypeCode ='002'，相同的'remainig'列和CodeName（也='1'）。

因此，下面给出的程序会生成这样的2个数据帧（df1和df2）然后生成结果，使它们的行“交织”。

import pandas as pd

data = [
    [ 'E123456', '102939485', 'Andrew',  'Hoover',  'hoovera@xyz.com',  '001', '002', 'AE', 'AWB', 'EMPLR' ],
    [ 'E123457', '675849302', 'Curt',    'Austin',  'austinc1@xyz.com', '001', '002', 'AE', 'AWB', 'EMPLR' ],
    [ 'E123458', '354852739', 'Celeste', 'Riddick', 'riddickc@xyz.com', '001', '002', 'AE', 'AWB', 'EMPLR' ],
    [ 'E123459', '937463528', 'Hazel',   'Tooley',  'tooleyh@xyz.com',  '001', '002', 'AE', 'AWB', 'EMPLR' ]
]
df = pd.DataFrame(data=data, columns=['employeeId', 'cumbId', 'firstName', 'lastName',
    'emailAddress', 'employeeIdTypeCode', 'cumbIDTypeCode', 'entityCode', 'sourceCode',
    'roleCode' ])
# 'Remainig' columns
cols = ['firstName', 'lastName', 'emailAddress', 'entityCode', 'sourceCode', 'roleCode']
# df1: employeeId, IDTypeCode = '001' and 'remainig' columns
df1 = df[['employeeId']].set_axis(['idvalue'], axis=1, inplace=False)
df1['IDTypeCode'] = '001'
df1 = df1.join(df[cols])
df1['CodeName'] = '1'
# df2: cumbId, IDTypeCode = '002' and 'remainig' columns
df2 = df[['cumbId']].set_axis(['idvalue'], axis=1, inplace=False)
df2['IDTypeCode'] = '002'
df2 = df2.join(df[cols])
df2['CodeName'] = '1'
# Result
result = pd.concat([df1,df2]).sort_index().reset_index(drop=True)

根据列名称转换熊猫中的数据框

2 个答案:

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